Question: Let $h(x)=\sqrt{\tan(e^x)}$. Find $h'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{\sqrt{\sec^2(e^x)}}$ (Choice B) B $\dfrac{e^x \sec^2(e^x)}{2\sqrt{\tan(e^x)}}$ (Choice C) C $\dfrac{ 1}{2\sqrt{\tan(e^x)}}$ (Choice D) D $\dfrac{ \sec^2(e^x)}{2\sqrt{\tan(e^x)}}$
Explanation: $h$ is a composition of three functions! Let... $u(x)=\sqrt{x}$ $v(x)=\tan(x)$ $w(x)=e^x$... then $h(x)=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $h'(x)$, we will need to use the chain rule twice! $\begin{aligned} h'(x)&=\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=\dfrac{1}{2\sqrt{x}}$ $v'(x)=\sec^2(x)$ $w'(x)=e^x$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={\dfrac{1}{2\sqrt{\tan(e^x)}}}\cdot{ \sec^2(e^x)}\cdot{e^x} \\\\ &=\dfrac{e^x\sec^2(e^x)}{2\sqrt{\tan(e^x)}} \end{aligned}$ In conclusion, $h'(x)=\dfrac{e^x\sec^2(e^x)}{2\sqrt{\tan(e^x)}}$.